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myininaya

5 years ago

Hey, so from that link the only thing I don't understand so far is why all of that shows that they are relatively prime. I do understand everything else though....
So still thinking as to why it shows they are....

anonymous

5 years ago

I thinks it has to do with the fact that every Fermat number is odd so can not be divided by two and hence the only gcd is 1

So when one have the recursion formula \[F _{n+1}=F _{0}*F _{1}*F _{2}*...*F _{n}+2\]
The solution is to show that the recursion formula is correct using induction, right?

myininaya

5 years ago

Yes :)

myininaya

5 years ago

They are proving it :)
Proving that will prove gcd(Fm,Fn)=1

myininaya

5 years ago

If you need help with the algebra, I can do that.
Right now, I'm multitasking though...so I will help slowly.

anonymous

5 years ago

So there are two things I don't understand the first is:\[F _{k+1}=2^{2^{k+1}}+1\]
\[=\left( \left( 2^{2^{k}} \right)^{2} +2*2^{2^{k}}+1 \right) - 2*2^{2^{k}}\]

anonymous

5 years ago

The second thing is \[\gcd(F _{m},F _{n})=\gcd(F _{m},2)\]

myininaya

5 years ago

\[=2^{2^{k+1}}+1=2^{2^k2^1}+1=2^{2^k 2}+1=(2^{2^k})^2+1\]

myininaya

5 years ago

They added in a zero.

anonymous

5 years ago

Why did the add the zero for?

myininaya

5 years ago

\[=(2^{2^k}+1)^2+1-2 \cdot 2^{2^k} \]
So they can write that one part as Fk

myininaya

5 years ago

or F_0F_1F_2...F_(K-1)+2

myininaya

5 years ago

They were setting up for the induction part

myininaya

5 years ago

\[=(2^{2^k}+1)(2^{2^k}+1)-2^{2^k}\]
Ignore that 1oops

myininaya

5 years ago

the one in expression i wrote before this one

myininaya

5 years ago

We are trying to show F_0F_1....F_k+2

anonymous

5 years ago

Ok I think I get the whole induction proof now.

anonymous

5 years ago

So seeing that \[F _{k+1}=F _{0}*F _{1}...*F _{k-1}*F _{k}+2\]
How do I know that \[GCD(F _{m},F _{n})=GCD(F _{m},2)\]

myininaya

5 years ago

Please ask any questions if you have them. I like this proof.

anonymous

5 years ago

* given that m

myininaya

5 years ago

Well gcd(odd,2)=1 since odd aren't divisible by 2

anonymous

5 years ago

Sorry formulated it bad, I mean how do I know that gcd F_n is 2 ?

myininaya

5 years ago

no Fn is not equal to 2

anonymous

5 years ago

* F_n is replaces by two *

anonymous

5 years ago

So where does the replacement of F_n to 2 come from?

myininaya

5 years ago

gcd(Fm,Fn)
=
gcd(Fm,F0F1...F(n-1)+2)
What about this?

anonymous

5 years ago

Sorry don't understand, Isn't that equal to f_(n+1)?

anonymous

5 years ago

Has it to do with the initial lead F_m | (F_n - 2) ?

myininaya

5 years ago

Well F_(n+1) equals
F0F1....Fn+2
----
Anyways,
Fm|(Fn-2)
=>for some integer k we have k*Fm=Fn-2
=> k*Fm+2=Fn
So gcd(Fm,Fn)
= gcd(Fm, k*Fm+2)
Let gcd(Fm, k*Fm+2)=d
=> for some integers a and b we have
aFm+b(k*Fm+2)=d
Fm(a+bk)+2b=d
But a+bk and b are just integers
We know gcd(Fm,2)=1 since Fm is odd and isn't divisible by 2.
This implies for some integers s and t we have sFm+2t=1
where s above is a+bk and t is b from above.
Did this make it more confusing?

myininaya

5 years ago

Implies d=1

anonymous

5 years ago

Not at all, that made perfect sense actually

anonymous

5 years ago

Will have to put it all together and solve it 2-3 times just to make sure it stays in my brain but you're given me the broad understanding, can't thank you enough!